Question: Evaluate the triple integral. $ \int_{-3}^1 \int_1^2 \int_0^4 y^2 - xz \, dy \, dx \, dz =$ Choose 1 answer: Choose 1 answer: (Choice A) A $\dfrac{141}{2}$ (Choice B) B $\dfrac{361}{3}$ (Choice C) C $\dfrac{328}{3}$ (Choice D) D $\dfrac{219}{2}$
We can evaluate triple integrals by repeated integration: $ \int_{a_0}^{a_1} \int_{b_0}^{b_1} \int_{c_0}^{c_1} f(x, y, z) \, dx \, dy \, dz = \int_{a_0}^{a_1} \left( \int_{b_0}^{b_1} \left[ \int_{c_0}^{c_1} f(x, y, z) \, dx \right] dy \right) dz$ The first layer: $\begin{aligned} &\int_{-3}^1 \int_1^2 \int_0^4 y^2 - xz \, dy \, dx \, dz \\ \\ &= \int_{-3}^1 \int_1^2 \left[ \dfrac{y^3}{3} - xzy \right]_0^4 dx \, dz \\ \\ &= \int_{-3}^1 \int_1^2 \dfrac{64}{3} - 4xz \, dx \, dz \end{aligned}$ The second layer: $\begin{aligned} &\int_{-3}^1 \int_1^2 \dfrac{64}{3} - 4xz \, dx \, dz \\ \\ &= \int_{-3}^1 \left[ \dfrac{64x}{3} - 2x^2z \right]_1^2 dz \\ \\ &= \int_{-3}^1 \left( \dfrac{128}{3} - 8z \right) - \left( \dfrac{64}{3} - 2z \right) dz \\ \\ &= \int_{-3}^1 \dfrac{64}{3} - 6z \, dz \end{aligned}$ The third layer: $\begin{aligned} &\int_{-3}^1 \dfrac{64}{3} - 6z \, dz \\ \\ &= \left[ \dfrac{64z}{3} - 3z^2 \right]_{-3}^1 \\ \\ &= \left( \dfrac{64}{3} - 3 \right) - \left( -64 - 27 \right) \\ \\ &= \dfrac{64}{3} + 88 \\ \\ &= \dfrac{328}{3} \end{aligned}$ In conclusion: $ \int_{-3}^1 \int_1^2 \int_0^4 y^2 - xz \, dy \, dx \, dz = \dfrac{328}{3}$